July 14, 2013

## How do you derive the Covariance equation?

How do you derive the equation Cov(x1, x2) = E[(x1-mu1)*(x2-mu2)] from the alternate equation for Covariance, Cov(x1,x2) = E(x1x2) – E(x1)E(x2) ? I am positive that I have to go from the latter to the first mentioned. I know E(x1)=mu1 and E(x2)=mu2 but I am not quite sure where to go from there.

Go the other way.

E[(x1-mu1)*(x2-mu2)] =
E[x1*x2-x1*mu2-x2*mu1+mu1*mu2] =
E[x1*x2] – E[x1*mu2] – E[x2*mu1] + mu1*mu2 =
E[x1*x2] – mu2*E[x1] – mu1*E[x2] + mu1*mu2 =
E[x1*x2] – mu2*mu1 – mu1*mu2 + mu1*mu2 =
E[x1*x2] – mu1*mu2=
E(x1*x2) – E(x1)*E(x2)

## What is the probability statement and how do you know?

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. The middle 30% of IQs fall between what two values?

Write the probability statement.
P(x1 < X < x2) = ____________________

Note that Z= (x – mu) /sigma is Gaussian distribution ( with mean 0 and standard deviation 1)

## How to solve for minor axis vertices for the ellipse?

Find the minor axis vertices for the ellipse. Write your answer in this form: (x1,y1),(x2,y2). Seperate the ordered pairs with a comma.
Equation= 9x^2+y^2+36x-4y+31=0
Center= (-2,2)
this is what i got so far I don’t know where to go from here.

Subtract 31 from both sides of the equation, getting:

9x^2 + 36x + y^2 – 4y = -31

9 divides 9x^2 and 36x, so we get

9*(x^2 + 4x) + (y^2 – 4y) = -31

Complete the squares adding the constants produced on the LHS to the RHS.

9*(x+2)^2 + (y-2)^2 = -31 + 36 + 4 = 40 – 31

9 * (x+2)^2 + (y-2)^2 = 9…..Divide everything by 9 and turn the denominators into squares.

(x+2)^2 . . . (y – 2)^2
———— + ————–
. 1^2 . . . . . . .3^2

Since 1 < 3 the minor axis is the center with first subtracting 1 from the -2
and then adding 1 from -2 while y is held constant at 2

(-2,2) + (1,0) = (-3,2)

(-2,2) + (1,0) = (-1,2)

The vertices of the minor axis are (-3,2) and (-1, 2)

Similarly hold the x (in the center) constant at -2 and first add and then
subtract 3 from the y = 2 to get the vertices of the major axis which
are therefore (-2,5) and (-2,-1).

.

## How do convert second order ODE to a system?

I have a second order ODE, I need to convert it to a system of first orders:
(a) mk”+kx’=0

(b) a system of second order, need a system of 4 first order ODE’s

m1x1”=-k1x1+k2(x2-x1)
m2x2”=-k2(x2-x1)

the #’s are just subscripts
any help would be appreciated.

Let X1 = x1, X2 = x2, X3 = x1′, X4 = x2′
and let Y = {X1,X2,X3,X4}

then Y’ =

X1′ = X3
X2′ = X4
X3′ = (1/m1) [ -k1 X1 + k2 (X2 - X1)
X4' = (1/m2) [ -k2 (X2 - X1) ]

If Y’ = A Y, then

A =
[ 0, 0, 1, 0]
[0, 0, 0, 1]
[(-k1/m1) + (-k2/m1), (k2/m1), 0, 0]
[(k2/m2), (-k2/m2), 0, 0]

Y’ = A Y is a system of 4 first order ODEs.